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{{task|Classic CS problems and programs}} [[Category:Simple]]
;Task: Implement the algorithm to compute the principal [[wp:Nth root|''n''th root]] of a positive real number ''A'', as explained at the [[wp:Nth root algorithm|Wikipedia page]].
360 Assembly
An example of converting integer floating-point using unnormalized short format. The 'include' file FORMAT, to format a floating point number, can be found in: [[360_Assembly_include|Include files 360 Assembly]].
* Nth root - x**(1/n) - 29/07/2018
NTHROOT CSECT
USING NTHROOT,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
SAVE (14,12) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
BAL R14,ROOTN call rootn(x,n)
LE F0,XN xn=rootn(x,n)
LA R0,6 decimals=6
BAL R14,FORMATF edit xn
MVC PG(13),0(R1) output xn
XPRNT PG,L'PG print buffer
L R13,4(0,R13) restore previous savearea pointer
RETURN (14,12),RC=0 restore registers from calling sav
ROOTN MVC ZN,=E'0' zn=0 ----------------------------
MVC ZN,N n
MVI ZN,X'46' zn=unnormalize(n)
LE F0,ZN zn
AE F0,=E'0' normalized
STE F0,ZN zn=normalize(n)
LE F6,=E'0' xm=0
LE F0,X x
DE F0,ZN /zn
STE F0,XN xn=x/zn
WHILEA LE F0,XN xn
SER F0,F6 xn-xm
LPER F0,F0 abs((xn-xm)
DE F0,XN /xn
CE F0,EPSILON while abs((xn-xm)/xn)>epsilon
BNH EWHILEA ~
LE F6,XN xm=xn
LE F0,ZN zn
SE F0,=E'1' zn-1
MER F0,F6 f0=(zn-1)*xm
L R2,N n
BCTR R2,0 n-1
LE F2,=E'1' xm
POW MER F2,F6 *xm
BCT R2,POW f2=xm**(n-1)
LE F4,X x
DER F4,F2 x/xm**(n-1)
AER F0,F4 (zn-1)*xm+x/xm**(n-1)
DE F0,ZN /zn
STE F0,XN xn=((zn-1)*xm+x/xm**(n-1))/zn
B WHILEA endwhile
EWHILEA LE F0,XN xn
BR R14 return ---------------------------
COPY FORMATF format a float
X DC E'2' x <== input
N DC F'2' n <== input
EPSILON DC E'1E-6' imprecision
XN DS E xn :: output
ZN DS E zn=float(n)
PG DC CL80' ' buffer
REGEQU
END NTHROOT
{{out}}
1.414213
Ada
The implementation is generic and supposed to work with any floating-point type. There is no result accuracy argument of Nth_Root, because the iteration is supposed to be monotonically descending to the root when starts at ''A''. Thus it should converge when this condition gets violated, i.e. when ''x''''k''+1''≥''x''''k''''.
with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Nth_Root is
generic
type Real is digits <>;
function Nth_Root (Value : Real; N : Positive) return Real;
function Nth_Root (Value : Real; N : Positive) return Real is
type Index is mod 2;
X : array (Index) of Real := (Value, Value);
K : Index := 0;
begin
loop
X (K + 1) := ( (Real (N) - 1.0) * X (K) + Value / X (K) ** (N-1) ) / Real (N);
exit when X (K + 1) >= X (K);
K := K + 1;
end loop;
return X (K + 1);
end Nth_Root;
function Long_Nth_Root is new Nth_Root (Long_Float);
begin
Put_Line ("1024.0 10th =" & Long_Float'Image (Long_Nth_Root (1024.0, 10)));
Put_Line (" 27.0 3rd =" & Long_Float'Image (Long_Nth_Root (27.0, 3)));
Put_Line (" 2.0 2nd =" & Long_Float'Image (Long_Nth_Root (2.0, 2)));
Put_Line ("5642.0 125th =" & Long_Float'Image (Long_Nth_Root (5642.0, 125)));
end Test_Nth_Root;
Sample output:
1024.0 10th = 2.00000000000000E+00
27.0 3rd = 3.00000000000000E+00
2.0 2nd = 1.41421356237310E+00
5642.0 125th = 1.07154759194477E+00
ALGOL 68
{{trans|C}}
{{works with|ALGOL 68|Standard - no extensions to language used}}
{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}} {{wont work with|ELLA ALGOL 68|Any (with appropriate job cards) - missing transput, and missing extended precision}}
REAL default p = 0.001;
PROC nth root = (INT n, LONG REAL a, p)LONG REAL:
(
[2]LONG REAL x := (a, a/n);
WHILE ABS(x[2] - x[1]) > p DO
x := (x[2], ((n-1)*x[2] + a/x[2]**(n-1))/n )
OD;
x[2]
);
PRIO ROOT = 8;
OP ROOT = (INT n, LONG REAL a)LONG REAL: nth root(n, a, default p);
OP ROOT = (INT n, INT a)LONG REAL: nth root(n, a, default p);
main:
(
printf(($2(" "gl)$,
nth root(10, LONG 7131.5 ** 10, default p),
nth root(5, 34, default p)));
printf(($2(" "gl)$,
10 ROOT ( LONG 7131.5 ** 10 ),
5 ROOT 34))
)
Output:
+7.131500000000000000001144390e +3
+2.024397462171090138953733623e +0
+7.131500000000000000001144390e +3
+2.024397462171090138953733623e +0
ALGOL W
begin
% nth root algorithm %
% returns the nth root of A, A must be > 0 %
% the required precision should be specified in precision %
long real procedure nthRoot( long real value A
; integer value n
; long real value precision
) ;
begin
long real xk, xd;
integer n1;
n1 := n - 1;
xk := A / n;
while begin
xd := ( ( A / ( xk ** n1 ) ) - xk ) / n;
xk := xk + xd;
abs( xd ) > precision
end do begin end;
xk
end nthRoot ;
% test cases %
r_format := "A"; r_w := 15; r_d := 6; % set output format %
write( nthRoot( 7131.5 ** 10, 10, 1'-5 ) );
write( nthRoot( 64, 6, 1'-5 ) );
end.
{{out}}
7131.500000
2.000000
ARM Assembly
{{works with|as|Raspberry Pi}}
/* ARM assembly Raspberry PI */
/* program nroot.s */
/* compile with option -mfpu=vfpv3 -mfloat-abi=hard */
/* link with gcc. Use C function for display float */
/* Constantes */
.equ EXIT, 1 @ Linux syscall
/* Initialized data */
.data
szFormat1: .asciz " %+09.15f\n"
.align 4
iNumberA: .int 1024
/* UnInitialized data */
.bss
.align 4
/* code section */
.text
.global main
main: @ entry of program
push {fp,lr} @ saves registers
/* root 10ieme de 1024 */
ldr r0,iAdriNumberA @ number address
ldr r0,[r0]
vmov s0,r0 @
vcvt.f64.s32 d0, s0 @conversion in float single précision (32 bits)
mov r0,#10 @ N
bl nthRoot
ldr r0,iAdrszFormat1 @ format
vmov r2,r3,d0
bl printf @ call C function !!!
@ Attention register dn lost !!!
/* square root of 2 */
vmov.f64 d1,#2.0 @ conversion 2 in float register d1
mov r0,#2 @ N
bl nthRoot
ldr r0,iAdrszFormat1 @ format
vmov r2,r3,d0
bl printf @ call C function !!!
100: @ standard end of the program
mov r0, #0 @ return code
pop {fp,lr} @restaur registers
mov r7, #EXIT @ request to exit program
swi 0 @ perform the system call
iAdrszFormat1: .int szFormat1
iAdriNumberA: .int iNumberA
/******************************************************************/
/* compute nth root */
/******************************************************************/
/* r0 contains N */
/* d0 contains the value */
/* d0 return result */
nthRoot:
push {r1,r2,lr} @ save registers
vpush {d1-d8} @ save float registers
FMRX r1,FPSCR @ copy FPSCR into r1
BIC r1,r1,#0x00370000 @ clears STRIDE and LEN
FMXR FPSCR,r1 @ copy r1 back into FPSCR
vmov s2,r0 @
vcvt.f64.s32 d6, s2 @ N conversion in float double précision (64 bits)
sub r1,r0,#1 @ N - 1
vmov s8,r1 @
vcvt.f64.s32 d4, s8 @conversion in float double précision (64 bits)
vmov.f64 d2,d0 @ a = A
vdiv.F64 d3,d0,d6 @ b = A/n
adr r2,dfPrec @ load précision
vldr d8,[r2]
1: @ begin loop
vmov.f64 d2,d3 @ a <- b
vmul.f64 d5,d3,d4 @ (N-1)*b
vmov.f64 d1,#1.0 @ constante 1 -> float
mov r2,#0 @ loop indice
2: @ compute pow (n-1)
vmul.f64 d1,d1,d3 @
add r2,#1
cmp r2,r1 @ n -1 ?
blt 2b @ no -> loop
vdiv.f64 d7,d0,d1 @ A / b pow (n-1)
vadd.f64 d7,d7,d5 @ + (N-1)*b
vdiv.f64 d3,d7,d6 @ / N -> new b
vsub.f64 d1,d3,d2 @ compute gap
vabs.f64 d1,d1 @ absolute value
vcmp.f64 d1,d8 @ compare float maj FPSCR
fmstat @ transfert FPSCR -> APSR
@ or use VMRS APSR_nzcv, FPSCR
bgt 1b @ if gap > précision -> loop
vmov.f64 d0,d3 @ end return result in d0
100:
vpop {d1-d8} @ restaur float registers
pop {r1,r2,lr} @ restaur arm registers
bx lr
dfPrec: .double 0f1E-10 @ précision
AutoHotkey
p := 0.000001
MsgBox, % nthRoot( 10, 7131.5**10, p) "`n"
. nthRoot( 5, 34.0 , p) "`n"
. nthRoot( 2, 2 , p) "`n"
. nthRoot(0.5, 7 , p) "`n"
;---------------------------------------------------------------------------
nthRoot(n, A, p) { ; http://en.wikipedia.org/wiki/Nth_root_algorithm
;---------------------------------------------------------------------------
x1 := A
x2 := A / n
While Abs(x1 - x2) > p {
x1 := x2
x2 := ((n-1)*x2+A/x2**(n-1))/n
}
Return, x2
}
Message box shows:
7131.500000
2.024397
1.414214
49.000000
AutoIt
;AutoIt Version: 3.2.10.0
$A=4913
$n=3
$x=20
ConsoleWrite ($n& " root of "& $A & " is " &nth_root_it($A,$n,$x))
ConsoleWrite ($n& " root of "& $A & " is " &nth_root_rec($A,$n,$x))
;Iterative
Func nth_root_it($A,$n,$x)
$x0="0"
While StringCompare(string($x0),string($x))
ConsoleWrite ($x&@CRLF)
$x0=$x
$x=((($n-1)*$x)+($A/$x^($n-1)))/$n
WEnd
Return $x
EndFunc
;Recursive
Func nth_root_rec($A,$n,$x)
ConsoleWrite ($x&@CRLF)
If $x==((($n-1)*$x)+($A/$x^($n-1)))/$n Then
Return $x
EndIf
Return nth_root_rec($A,$n,((($n-1)*$x)+($A/$x^($n-1)))/$n)
EndFunc
output :
20
17.4275
17.0104009124137
17.0000063582823
17.0000000000024
17
3 root of 4913 is 17
AWK
#!/usr/bin/awk -f
BEGIN {
# test
print nthroot(8,3)
print nthroot(16,2)
print nthroot(16,4)
print nthroot(125,3)
print nthroot(3,3)
print nthroot(3,2)
}
function nthroot(y,n) {
eps = 1e-15; # relative accuracy
x = 1;
do {
d = ( y / ( x^(n-1) ) - x ) / n ;
x += d;
e = eps*x; # absolute accuracy
} while ( d < -e || d > e )
return x
}
Sample output: 2 4 2 5 1.44225 1.73205
BASIC
{{works with|QBasic}} {{works with|FreeBASIC}} {{works with|PowerBASIC}} {{works with|Visual Basic}}
This function is fairly generic MS BASIC. It could likely be used in most modern BASICs with little or no change.
FUNCTION RootX (tBase AS DOUBLE, tExp AS DOUBLE, diffLimit AS DOUBLE) AS DOUBLE
DIM tmp1 AS DOUBLE, tmp2 AS DOUBLE
' Initial guess:
tmp1 = tBase / tExp
DO
tmp2 = tmp1
' 1# tells compiler that "1" is a double, not an integer
tmp1 = (((tExp - 1#) * tmp2) + (tBase / (tmp2 ^ (tExp - 1#)))) / tExp
LOOP WHILE (ABS(tmp1 - tmp2) > diffLimit)
RootX = tmp1
END FUNCTION
Note that for the above to work in QBasic, the function definition needs to be changed like so:
FUNCTION RootX# (tBase AS DOUBLE, tExp AS DOUBLE, diffLimit AS DOUBLE)
The function is called like so:
PRINT "The "; e; "th root of "; b; " is "; RootX(b, e, .000001)
Sample output: The 4th root of 16 is 2
For BASICs without the '''^''' operator, it would be trivial to write a function to reproduce it (as is done in the [[#C|C]] example below).
See also the [[#Liberty BASIC|Liberty BASIC]] and [[#PureBasic|PureBasic]] solutions.
BBC BASIC
{{works with|BBC BASIC for Windows}}
*FLOAT 64
@% = &D0D
PRINT "Cube root of 5 is "; FNroot(3, 5, 0)
PRINT "125th root of 5643 is "; FNroot(125, 5643, 0)
END
DEF FNroot(n%, a, d)
LOCAL x0, x1 : x0 = a / n% : REM Initial guess
REPEAT
x1 = ((n% - 1)*x0 + a/x0^(n%-1)) / n%
SWAP x0, x1
UNTIL ABS (x0 - x1) <= d
= x0
'''Output:'''
Cube root of 5 is 1.709975946677
125th root of 5643 is 1.071549111198
bc
/* Take the nth root of 'a' (a positive real number).
* 'n' must be an integer.
* Result will have 'd' digits after the decimal point.
*/
define r(a, n, d) {
auto e, o, x, y, z
if (n == 0) return(1)
if (a == 0) return(0)
o = scale
scale = d
e = 1 / 10 ^ d
if (n < 0) {
n = -n
a = 1 / a
}
x = 1
while (1) {
y = ((n - 1) * x + a / x ^ (n - 1)) / n
z = x - y
if (z < 0) z = -z
if (z < e) break
x = y
}
scale = o
return(y)
}
Bracmat
Bracmat does not have floating point numbers as primitive type. Instead we have to use rational numbers. This code is not fast!
( ( root
= n a d x0 x1 d2 rnd 10-d
. ( rnd { For 'rounding' rational numbers = keep number of digits within bounds. }
= N r
. !arg:(?N.?r)
& div$(!N*!r+1/2.1)*!r^-1
)
& !arg:(?n,?a,?d)
& !a*!n^-1:?x0
& 10^(-1*!d):?10-d
& whl
' ( ( rnd$(((!n+-1)*!x0+!a*!x0^(1+-1*!n))*!n^-1.10^!d)
. !x0
)
: (?x0.?x1)
& (!x0+-1*!x1)^2:~<!10-d { Exit loop when required precision is reached. }
)
& flt$(!x0,!d) { Convert rational number to floating point representation. }
)
& ( show
= N A precision
. !arg:(?N,?A,?precision)
& out$(str$(!A "^(" !N^-1 ")=" root$(!N,!A,!precision)))
)
& show$(10,1024,20)
& show$(3,27,20)
& show$(2,2,100)
& show$(125,5642,20)
)
Output:
1024^(1/10)=2,00000000000000000000*10E0
27^(1/3)=3,00000000000000000000*10E0
2^(1/2)=1,4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727*10E0
5642^(1/125)=1,07154759194476751170*10E0
C
Implemented without using math library, because if we were to use pow(), the whole exercise wouldn't make sense.
#include <stdio.h>
#include <float.h>
double pow_ (double x, int e) {
int i;
double r = 1;
for (i = 0; i < e; i++) {
r *= x;
}
return r;
}
double root (int n, double x) {
double d, r = 1;
if (!x) {
return 0;
}
if (n < 1 || (x < 0 && !(n&1))) {
return 0.0 / 0.0; /* NaN */
}
do {
d = (x / pow_(r, n - 1) - r) / n;
r += d;
}
while (d >= DBL_EPSILON * 10 || d <= -DBL_EPSILON * 10);
return r;
}
int main () {
int n = 15;
double x = pow_(-3.14159, 15);
printf("root(%d, %g) = %g\n", n, x, root(n, x));
return 0;
}
C#
Almost exactly how C works.
static void Main(string[] args)
{
Console.WriteLine(NthRoot(81,2,.001));
Console.WriteLine(NthRoot(1000,3,.001));
Console.ReadLine();
}
public static double NthRoot(double A,int n, double p)
{
double _n= (double) n;
double[] x = new double[2];
x[0] = A;
x[1] = A/_n;
while(Math.Abs(x[0] -x[1] ) > p)
{
x[1] = x[0];
x[0] = (1/_n)*(((_n-1)*x[1]) + (A/Math.Pow(x[1],_n-1)));
}
return x[0];
}
C++
double NthRoot(double m_nValue, double index, double guess, double pc)
{
double result = guess;
double result_next;
do
{
result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0))));
result = result_next;
pc--;
}while(pc>1);
return result;
};
double NthRoot(double value, double degree)
{
return pow(value, (double)(1 / degree));
};
Clojure
(ns test-project-intellij.core
(:gen-class))
;; define abs & power to avoid needing to bring in the clojure Math library
(defn abs [x]
" Absolute value"
(if (< x 0) (- x) x))
(defn power [x n]
" x to power n, where n = 0, 1, 2, ... "
(apply * (repeat n x)))
(defn calc-delta [A x n]
" nth rooth algorithm delta calculation "
(/ (- (/ A (power x (- n 1))) x) n))
(defn nth-root
" nth root of algorithm: A = numer, n = root"
([A n] (nth-root A n 0.5 1.0)) ; Takes only two arguments A, n and calls version which takes A, n, guess-prev, guess-current
([A n guess-prev guess-current] ; version take takes in four arguments (A, n, guess-prev, guess-current)
(if (< (abs (- guess-prev guess-current)) 1e-6)
guess-current
(recur A n guess-current (+ guess-current (calc-delta A guess-current n)))))) ; iterate answer using tail recursion
CoffeeScript
nth_root = (A, n, precision=0.0000000000001) ->
x = 1
while true
x_new = (1 / n) * ((n - 1) * x + A / Math.pow(x, n - 1))
return x_new if Math.abs(x_new - x) < precision
x = x_new
# tests
do ->
tests = [
[8, 3]
[16, 4]
[32, 5]
[343, 3]
[1024, 10]
[1000000000, 3]
[1000000000, 9]
[100, 2]
[100, 3]
[100, 5]
[100, 10]
]
for test in tests
[x, n] = test
root = nth_root x, n
console.log "#{x} root #{n} = #{root} (root^#{n} = #{Math.pow root, n})"
output
coffee nth_root.coffee 8 root 3 = 2 (root^3 = 8) 16 root 4 = 2 (root^4 = 16) 32 root 5 = 2 (root^5 = 32) 343 root 3 = 7 (root^3 = 343) 1024 root 10 = 2 (root^10 = 1024) 1000000000 root 3 = 1000 (root^3 = 1000000000) 1000000000 root 9 = 10 (root^9 = 1000000000) 100 root 2 = 10 (root^2 = 100) 100 root 3 = 4.641588833612778 (root^3 = 99.99999999999997) 100 root 5 = 2.5118864315095806 (root^5 = 100.0000000000001) 100 root 10 = 1.5848931924611134 (root^10 = 99.99999999999993)
## Common Lisp
This version does not check for cycles in <var>x<sub>i</sub></var> and <var>x<sub>i+1</sub></var>, but finishes when the difference between them drops below <var>ε</var>. The initial guess can be provided, but defaults to <var>n-1</var>.
```lisp
(defun nth-root (n a &optional (epsilon .0001) (guess (1- n)))
(assert (and (> n 1) (> a 0)))
(flet ((next (x)
(/ (+ (* (1- n) x)
(/ a (expt x (1- n))))
n)))
(do* ((xi guess xi+1)
(xi+1 (next xi) (next xi)))
((< (abs (- xi+1 xi)) epsilon) xi+1))))
nth-root may return rationals rather than floating point numbers, so easy checking for correctness may require coercion to floats. For instance,
(let* ((r (nth-root 3 10))
(rf (coerce r 'float)))
(print (* r r r ))
(print (* rf rf rf)))
produces the following output.
1176549099958810982335712173626176/117654909634627320192156007194483
10.0
D
import std.stdio, std.math;
real nthroot(in int n, in real A, in real p=0.001) pure nothrow {
real[2] x = [A, A / n];
while (abs(x[1] - x[0]) > p)
x = [x[1], ((n - 1) * x[1] + A / (x[1] ^^ (n-1))) / n];
return x[1];
}
void main() {
writeln(nthroot(10, 7131.5 ^^ 10));
writeln(nthroot(6, 64));
}
{{out}}
7131.5
2
Delphi
USES
Math;
function NthRoot(A, Precision: Double; n: Integer): Double;
var
x_p, X: Double;
begin
x_p := Sqrt(A);
while Abs(A - Power(x_p, n)) > Precision do
begin
x := (1/n) * (((n-1) * x_p) + (A/(Power(x_p, n - 1))));
x_p := x;
end;
Result := x_p;
end;
E
Rather than choosing an arbitrary precision, this implementation continues until a cycle in the iterated result is found, thus producing an answer almost as precise as the number type.
(Disclaimer: This was not written by a numerics expert; there may be reasons this is a bad idea. Also, it might be that cycles are always of length 2, which would reduce the amount of calculation needed by 2/3.)
def nthroot(n, x) {
require(n > 1 && x > 0)
def np := n - 1
def iter(g) { return (np*g + x/g**np) / n }
var g1 := x
var g2 := iter(g1)
while (!(g1 <=> g2)) {
g1 := iter(g1)
g2 := iter(iter(g2))
}
return g1
}
EasyLang
## Elixir
{{trans|Erlang}}
```elixir
defmodule RC do
def nth_root(n, x, precision \\ 1.0e-5) do
f = fn(prev) -> ((n - 1) * prev + x / :math.pow(prev, (n-1))) / n end
fixed_point(f, x, precision, f.(x))
end
defp fixed_point(_, guess, tolerance, next) when abs(guess - next) < tolerance, do: next
defp fixed_point(f, _, tolerance, next), do: fixed_point(f, next, tolerance, f.(next))
end
Enum.each([{2, 2}, {4, 81}, {10, 1024}, {1/2, 7}], fn {n, x} ->
IO.puts "#{n} root of #{x} is #{RC.nth_root(n, x)}"
end)
{{out}}
2 root of 2 is 1.4142135623746899
4 root of 81 is 3.0
10 root of 1024 is 2.00000000022337
0.5 root of 7 is 48.99999999999993
Erlang
Done by finding the fixed point of a function, which aims to find a value of x for which f(x)=x:
fixed_point(F, Guess, Tolerance) ->
fixed_point(F, Guess, Tolerance, F(Guess)).
fixed_point(_, Guess, Tolerance, Next) when abs(Guess - Next) < Tolerance ->
Next;
fixed_point(F, _, Tolerance, Next) ->
fixed_point(F, Next, Tolerance, F(Next)).
The nth root function algorithm defined on the wikipedia page linked above can advantage of this:
nth_root(N, X) -> nth_root(N, X, 1.0e-5).
nth_root(N, X, Precision) ->
F = fun(Prev) -> ((N - 1) * Prev + X / math:pow(Prev, (N-1))) / N end,
fixed_point(F, X, Precision).
Excel
This will work in any spreadsheet that uses Excel-compatible expressions -- i.e. [[wp:KOffice|KOffice]]'s KCells (formerly KSpread), [[wp:Calligra Tables|Calligra Tables]], and [[wp:OpenOffice.org Calc|OpenOffice.org Calc]].
Beside the obvious;
=A1^(1/B1)
*Cell A1 is the base. *Cell B1 is the exponent. *Cell A2 is the first guess (any non-zero number will do). In cell A3, enter this formula: =((($B$1-1)*A2)+($A$1/(A2^($B$1-1))))/$B$1 Copy A3 down until you get 2 cells with the same value. (Once there are two visibly-identical cells, all cells below those two will also be identical.)
For example, here we calculate the cube root of 100: {| class="wikitable" |- | | '''A'''
| '''B''' |
|---|
| '''1''' |
| 100 |
| 3 |
| - |
| '''2''' |
| 7 |
| (first guess) |
| - |
| '''3''' |
| 5.346938776 |
| |- | '''4''' | 4.730544697 | |- | '''5''' | 4.643251125 | |- | '''6''' | 4.641589429 | |- | '''7''' | 4.641588834 | |- | '''8''' | 4.641588834 | |}
Alternately, Excel could use the [[#BASIC|BASIC]] example above as VBA code, deleting A2 and replacing A3's formula with something like this: =RootX(A1,B1,.00000001)
=={{header|F_Sharp|F#}}==
let nthroot n A =
let rec f x =
let m = n - 1.
let x' = (m * x + A/x**m) / n
match abs(x' - x) with
| t when t < abs(x * 1e-9) -> x'
| _ -> f x'
f (A / double n)
[<EntryPoint>]
let main args =
if args.Length <> 2 then
eprintfn "usage: nthroot n A"
exit 1
let (b, n) = System.Double.TryParse(args.[0])
let (b', A) = System.Double.TryParse(args.[1])
if (not b) || (not b') then
eprintfn "error: parameter must be a number"
exit 1
printf "%A" (nthroot n A)
0
Compiled using fsc nthroot.fs example output:
nthroot 0.5 7
49.0
Factor
{{trans|Forth}}
USING: kernel locals math math.functions prettyprint ;
:: th-root ( a n -- a^1/n )
a [
a over n 1 - ^ /f
over n 1 - *
+ n /f
swap over 1e-5 ~ not
] loop ;
34 5 th-root . ! 2.024397458499888
34 5 recip ^ . ! 2.024397458499888
Forth
: th-root { F: a F: n -- a^1/n }
a
begin
a fover n 1e f- f** f/
fover n 1e f- f*
f+ n f/
fswap fover 1e-5 f~
until ;
34e 5e th-root f. \ 2.02439745849989
34e 5e 1/f f** f. \ 2.02439745849989
Fortran
program NthRootTest
implicit none
print *, nthroot(10, 7131.5**10)
print *, nthroot(5, 34.0)
contains
function nthroot(n, A, p)
real :: nthroot
integer, intent(in) :: n
real, intent(in) :: A
real, intent(in), optional :: p
real :: rp, x(2)
if ( A < 0 ) then
stop "A < 0" ! we handle only real positive numbers
elseif ( A == 0 ) then
nthroot = 0
return
end if
if ( present(p) ) then
rp = p
else
rp = 0.001
end if
x(1) = A
x(2) = A/n ! starting "guessed" value...
do while ( abs(x(2) - x(1)) > rp )
x(1) = x(2)
x(2) = ((n-1.0)*x(2) + A/(x(2) ** (n-1.0)))/real(n)
end do
nthroot = x(2)
end function nthroot
end program NthRootTest
FreeBASIC
' version 14-01-2019
' compile with: fbc -s console
Function nth_root(n As Integer, number As Double) As Double
Dim As Double a1 = number / n, a2 , a3
Do
a3 = Abs(a2 - a1)
a2 = ((n -1) * a1 + number / a1 ^ (n -1)) / n
Swap a1, a2
Loop Until Abs(a2 - a1) = a3
Return a1
End Function
' ------=< MAIN >=------
Dim As UInteger n
Dim As Double tmp
Print
Print " n 5643 ^ 1 / n nth_root ^ n"
Print " ------------------------------------"
For n = 3 To 11 Step 2
tmp = nth_root(n, 5643)
Print Using " ### ###.######## ####.########"; n; tmp; tmp ^ n
Next
Print
For n = 25 To 125 Step 25
tmp = nth_root(n, 5643)
Print Using " ### ###.######## ####.########"; n; tmp; tmp ^ n
Next
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
n 5643 ^ 1 / n nth_root ^ n
------------------------------------
3 17.80341642 5643.00000000
5 5.62732516 5643.00000000
7 3.43502583 5643.00000000
9 2.61116581 5643.00000000
11 2.19303907 5643.00000000
25 1.41273402 5643.00000000
50 1.18858488 5643.00000000
75 1.12207047 5643.00000000
100 1.09022240 5643.00000000
125 1.07154911 5643.00000000
FutureBasic
include "ConsoleWindow"
def tab 8
local fn NthRoot( root as long, a as long, precision as double ) as double
dim as double x0, x1
x0 = a : x1 = a /root
while ( abs( x1 - x0 ) > precision )
x0 = x1
x1 = ( ( root -1.0 ) * x1 + a / x1 ^ ( root -1.0 ) ) /root
wend
end fn = x1
print " 125th Root of 5643 Precision .001", using "#.###############"; fn NthRoot( 125, 5642, 0.001 )
print " 125th Root of 5643 Precision .001", using "#.###############"; fn NthRoot( 125, 5642, 0.001 )
print " 125th Root of 5643 Precision .00001", using "#.###############"; fn NthRoot( 125, 5642, 0.00001 )
print " Cube Root of 27 Precision .00001", using "#.###############"; fn NthRoot( 3, 27, 0.00001 )
print "Square Root of 2 Precision .00001", using "#.###############"; fn NthRoot( 2, 2, 0.00001 )
print "Square Root of 2 Precision .00001", using "#.###############"; sqr(2) // Processor floating point calc deviation
print " 10th Root of 1024 Precision .00001", using "#.###############"; fn NthRoot( 10, 1024, 0.00001 )
print " 5th Root of 34 Precision .00001", using "#.###############"; fn NthRoot( 5, 34, 0.00001 )
Output:
125th Root of 5643 Precision .001 1.071559602191682
125th Root of 5643 Precision .001 1.071559602191682
125th Root of 5643 Precision .00001 1.071547591944772
Cube Root of 27 Precision .00001 3.000000000000002
Square Root of 2 Precision .00001 1.414213562374690
Square Root of 2 Precision .00001 1.414213562373095
10th Root of 1024 Precision .00001 2.000000000000000
5th Root of 34 Precision .00001 2.024397458499885
Go
func root(a float64, n int) float64 {
n1 := n - 1
n1f, rn := float64(n1), 1/float64(n)
x, x0 := 1., 0.
for {
potx, t2 := 1/x, a
for b := n1; b > 0; b >>= 1 {
if b&1 == 1 {
t2 *= potx
}
potx *= potx
}
x0, x = x, rn*(n1f*x+t2)
if math.Abs(x-x0)*1e15 < x {
break
}
}
return x
}
The above version is for 64 bit wide floating point numbers. The following uses math/big Float to implement this same function with 256 bits of precision.
''A set of wrapper functions around the somewhat muddled big math library functions is used to make the main function more readable, and also it was necessary to create a power function (Exp) as the library also lacks this function.'' '''The exponent in the limit must be at least one less than the number of bits of precision of the input value or the function will enter an infinite loop!'''
import "math/big"
func Root(a *big.Float, n uint64) *big.Float {
limit := Exp(New(2), 256)
n1 := n-1
n1f, rn := New(float64(n1)), Div(New(1.0), New(float64(n)))
x, x0 := New(1.0), Zero()
_ = x0
for {
potx, t2 := Div(New(1.0), x), a
for b:=n1; b>0; b>>=1 {
if b&1 == 1 {
t2 = Mul(t2, potx)
}
potx = Mul(potx, potx)
}
x0, x = x, Mul(rn, Add(Mul(n1f, x), t2) )
if Lesser(Mul(Abs(Sub(x, x0)), limit), x) { break }
}
return x
}
func Abs(a *big.Float) *big.Float {
return Zero().Abs(a)
}
func Exp(a *big.Float, e uint64) *big.Float {
result := Zero().Copy(a)
for i:=uint64(0); i<e-1; i++ {
result = Mul(result, a)
}
return result
}
func New(f float64) *big.Float {
r := big.NewFloat(f)
r.SetPrec(256)
return r
}
func Div(a, b *big.Float) *big.Float {
return Zero().Quo(a, b)
}
func Zero() *big.Float {
r := big.NewFloat(0.0)
r.SetPrec(256)
return r
}
func Mul(a, b *big.Float) *big.Float {
return Zero().Mul(a, b)
}
func Add(a, b *big.Float) *big.Float {
return Zero().Add(a, b)
}
func Sub(a, b *big.Float) *big.Float {
return Zero().Sub(a, b)
}
func Lesser(x, y *big.Float) bool {
return x.Cmp(y) == -1
}
Groovy
Solution:
import static Constants.tolerance
import static java.math.RoundingMode.HALF_UP
def root(double base, double n) {
double xOld = 1
double xNew = 0
while (true) {
xNew = ((n - 1) * xOld + base/(xOld)**(n - 1))/n
if ((xNew - xOld).abs() < tolerance) { break }
xOld = xNew
}
(xNew as BigDecimal).setScale(7, HALF_UP)
}
Test:
class Constants {
static final tolerance = 0.00001
}
print '''
Base Power Calc'd Root Actual Root
------- ------ ----------- -----------
'''
def testCases = [
[b:32.0, n:5.0, r:2.0],
[b:81.0, n:4.0, r:3.0],
[b:Math.PI**2, n:4.0, r:Math.PI**(0.5)],
[b:7.0, n:0.5, r:49.0],
]
testCases.each {
def r = root(it.b, it.n)
printf('%7.4f %6.4f %11.4f %11.4f\n',
it.b, it.n, r, it.r)
assert (r - it.r).abs() <= tolerance
}
Output:
Base Power Calc'd Root Actual Root
------- ------ ----------- -----------
32.0000 5.0000 2.0000 2.0000
81.0000 4.0000 3.0000 3.0000
9.8696 4.0000 1.7725 1.7725
7.0000 0.5000 49.0000 49.0000
Haskell
Function exits when there's no difference between two successive values.
n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/n)
Use:
*Main> 2 `nthRoot` 2
1.414213562373095
*Main> 5 `nthRoot` 34
2.024397458499885
*Main> 10 `nthRoot` (734^10)
734.0
*Main> 0.5 `nthRoot` 7
49.0
Or, in applicative terms, with formatted output:
nthRoot :: Double -> Double -> Double
nthRoot n x =
fst $
until
(uncurry (==))
(((,) <*> ((/ n) . ((+) <$> ((n - 1) *) <*> (x /) . (** (n - 1))))) . snd)
(x, x / n)
-- TESTS --------------------------------------------------
main :: IO ()
main =
putStrLn $
fTable
"Nth roots:"
(\(a, b) -> show a ++ " `nthRoot` " ++ show b)
show
(uncurry nthRoot)
[(2, 2), (5, 34), (10, 734 ^ 10), (0.5, 7)]
-- FORMAT OF RESULTS --------------------------------------
fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String
fTable s xShow fxShow f xs =
let w = maximum (length . xShow <$> xs)
rjust n c = drop . length <*> (replicate n c ++)
in unlines $
s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs
{{Out}}
Nth roots:
2.0 `nthRoot` 2.0 -> 1.414213562373095
5.0 `nthRoot` 34.0 -> 2.0243974584998847
10.0 `nthRoot` 4.539004352165717e28 -> 734.0
0.5 `nthRoot` 7.0 -> 49.0
HicEst
WRITE(Messagebox) NthRoot(5, 34)
WRITE(Messagebox) NthRoot(10, 7131.5^10)
FUNCTION NthRoot(n, A)
REAL :: prec = 0.001
IF( (n > 0) * (A > 0) ) THEN
NthRoot = A / n
DO i = 1, 1/prec
x = ((n-1)*NthRoot + A/(NthRoot^(n-1))) / n
IF( ABS(x - NthRoot) <= prec ) THEN
RETURN
ENDIF
NthRoot = x
ENDDO
ENDIF
WRITE(Messagebox, Name) 'Cannot solve problem for:', prec, n, A
END
=={{header|Icon}} and {{header|Unicon}}== All Icon/Unicon reals are double precision.
procedure main()
showroot(125,3)
showroot(27,3)
showroot(1024,10)
showroot(39.0625,4)
showroot(7131.5^10,10)
end
procedure showroot(a,n)
printf("%i-th root of %i = %i\n",n,a,root(a,n))
end
procedure root(a,n,p) #: finds the n-th root of the number a to precision p
if n < 0 | type(n) !== "integer" then runerr(101,n)
if a < 0 then runerr(205,a)
/p := 1e-14 # precision
xn := a / real(n) # initial guess
while abs(a - xn^n) > p do
xn := ((n - 1) * (xi := xn) + a / (xi ^ (n-1))) / real(n)
return xn
end
link printf
Output:
3-th root of 125 = 5.0
3-th root of 27 = 3.0
10-th root of 1024 = 2.0
4-th root of 39.0625 = 2.5
10-th root of 3.402584077894253e+038 = 7131.5
J
{{trans|E}}
J has a built in Nth root primitive, '''%:'''. For example, '''7131.5 = 10 %: 7131.5^10'''. Also, the exponentiation primitive supports exponents < 1, e.g. '''7131.5 = (7131.5^10)^(1%10)'''.
But, since the [[Talk:Nth_root_algorithm#Comparison_to_Non-integer_Exponentiation|talk page discourages]] using built-in facilities, here is a reimplementation, using the [[#E|E]] algorithm:
'`N X NP' =. (0 { [)`(1 { [)`(2 { [)
iter =. N %~ (NP * ]) + X % ] ^ NP
nth_root =: (, , _1+[) iter^:_ f. ]
10 nth_root 7131.5^10
7131.5
Java
{{trans|Fortran}}
public static double nthroot(int n, double A) {
return nthroot(n, A, .001);
}
public static double nthroot(int n, double A, double p) {
if(A < 0) {
System.err.println("A < 0");// we handle only real positive numbers
return -1;
} else if(A == 0) {
return 0;
}
double x_prev = A;
double x = A / n; // starting "guessed" value...
while(Math.abs(x - x_prev) > p) {
x_prev = x;
x = ((n - 1.0) * x + A / Math.pow(x, n - 1.0)) / n;
}
return x;
}
{{trans|E}}
public static double nthroot(int n, double x) {
assert (n > 1 && x > 0);
int np = n - 1;
double g1 = x;
double g2 = iter(g1, np, n, x);
while (g1 != g2) {
g1 = iter(g1, np, n, x);
g2 = iter(iter(g2, np, n, x), np, n, x);
}
return g1;
}
private static double iter(double g, int np, int n, double x) {
return (np * g + x / Math.pow(g, np)) / n;
}
JavaScript
Gives the ''n'':nth root of ''num'', with precision ''prec''. (''n'' defaults to 2 [e.g. sqrt], ''prec'' defaults to 12.)
function nthRoot(num, nArg, precArg) {
var n = nArg || 2;
var prec = precArg || 12;
var x = 1; // Initial guess.
for (var i=0; i<prec; i++) {
x = 1/n * ((n-1)*x + (num / Math.pow(x, n-1)));
}
return x;
}
jq
# An iterative algorithm for finding: self ^ (1/n) to the given
# absolute precision if "precision" > 0, or to within the precision
# allowed by IEEE 754 64-bit numbers.
# The following implementation handles underflow caused by poor estimates.
def iterative_nth_root(n; precision):
def abs: if . < 0 then -. else . end;
def sq: .*.;
def pow(p): . as $in | reduce range(0;p) as $i (1; . * $in);
def _iterate: # state: [A, x1, x2, prevdelta]
.[0] as $A | .[1] as $x1 | .[2] as $x2 | .[3] as $prevdelta
| ( $x2 | pow(n-1)) as $power
| if $power <= 2.155094094640383e-309
then [$A, $x1, ($x1 + $x2)/2, n] | _iterate
else (((n-1)*$x2 + ($A/$power))/n) as $x1
| (($x1 - $x2)|abs) as $delta
| if (precision == 0 and $delta == $prevdelta and $delta < 1e-15)
or (precision > 0 and $delta <= precision) or $delta == 0 then $x1
else [$A, $x2, $x1, $delta] | _iterate
end
end
;
if n == 1 then .
elif . == 0 then 0
elif . < 0 then error("iterative_nth_root: input \(.) < 0")
elif n != (n|floor) then error("iterative_nth_root: argument \(n) is not an integer")
elif n == 0 then error("iterative_nth_root(0): domain error")
elif n < 0 then 1/iterative_nth_root(-n; precision)
else [., ., (./n), n, 0] | _iterate
end
;
'''Example''': Compare the results of iterative_nth_root and nth_root implemented using builtins
def demo(x):
def nth_root(n): log / n | exp;
def lpad(n): tostring | (n - length) * " " + .;
. as $in
| "\(x)^(1/\(lpad(5))): \(x|nth_root($in)|lpad(18)) vs \(x|iterative_nth_root($in; 1e-10)|lpad(18)) vs \(x|iterative_nth_root($in; 0))"
;
# 5^m for various values of n:
"5^(1/ n): builtin precision=1e-10 precision=0",
( (1,-5,-3,-1,1,3,5,1000,10000) | demo(5))
{{Out}}
$ jq -n -r -f nth_root_machine_precision.jq
5^(1/ n): builtin precision=1e-10 precision=0
5^(1/ 1): 4.999999999999999 vs 5 vs 5
5^(1/ -5): 0.7247796636776955 vs 0.7247796636776956 vs 0.7247796636776955
5^(1/ -3): 0.5848035476425733 vs 0.5848035476425731 vs 0.5848035476425731
5^(1/ -1): 0.2 vs 0.2 vs 0.2
5^(1/ 1): 4.999999999999999 vs 5 vs 5
5^(1/ 3): 1.709975946676697 vs 1.709975946676697 vs 1.709975946676697
5^(1/ 5): 1.3797296614612147 vs 1.3797296614612147 vs 1.379729661461215
5^(1/ 1000): 1.0016107337527294 vs 1.0016107337527294 vs 1.0016107337527294
5^(1/10000): 1.0001609567433902 vs 1.0001609567433902 vs 1.0001609567433902
Julia
{{works with|Julia|1.2}}
Julia has a built-in exponentiation function A^(1 / n), but the specification calls for us to use Newton's method (which we iterate until the limits of machine precision are reached):
function nthroot(n::Integer, r::Real)
r < 0 || n == 0 && throw(DomainError())
n < 0 && return 1 / nthroot(-n, r)
r > 0 || return 0
x = r / n
prevdx = r
while true
y = x ^ (n - 1)
dx = (r - y * x) / (n * y)
abs(dx) ≥ abs(prevdx) && return x
x += dx
prevdx = dx
end
end
@show nthroot.(-5:2:5, 5.0)
@show nthroot.(-5:2:5, 5.0) - 5.0 .^ (1 ./ (-5:2:5))
{{out}}
nthroot.(-5:2:5, 5.0) = [0.7247796636776955, 0.5848035476425731, 0.2, 5.0, 1.709975946676697, 1.379729661461215]
nthroot.(-5:2:5, 5.0) - 5.0 .^ (1 ./ (-5:2:5)) = [0.0, -1.1102230246251565e-16, 0.0, 0.0, 0.0, 0.0]
Kotlin
{{trans|E}}
// version 1.0.6
fun nthRoot(x: Double, n: Int): Double {
if (n < 2) throw IllegalArgumentException("n must be more than 1")
if (x <= 0.0) throw IllegalArgumentException("x must be positive")
val np = n - 1
fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n
var g1 = x
var g2 = iter(g1)
while (g1 != g2) {
g1 = iter(g1)
g2 = iter(iter(g2))
}
return g1
}
fun main(args: Array<String>) {
val numbers = arrayOf(1728.0 to 3, 1024.0 to 10, 2.0 to 2)
for (number in numbers)
println("${number.first} ^ 1/${number.second}\t = ${nthRoot(number.first, number.second)}")
}
{{out}}
1728.0 ^ 1/3 = 12.0
1024.0 ^ 1/10 = 2.0
2.0 ^ 1/2 = 1.414213562373095
Liberty BASIC
print "First estimate is: ", using( "#.###############", NthRoot( 125, 5642, 0.001 ));
print " ... and better is: ", using( "#.###############", NthRoot( 125, 5642, 0.00001))
print "125'th root of 5642 by LB's exponentiation operator is "; using( "#.###############", 5642^(1 /125))
print "27^(1 / 3)", using( "#.###############", NthRoot( 3, 27, 0.00001))
print "2^(1 / 2)", using( "#.###############", NthRoot( 2, 2, 0.00001))
print "1024^(1 /10)", using( "#.###############", NthRoot( 10, 1024, 0.00001))
wait
function NthRoot( n, A, p)
x( 0) =A
x( 1) =A /n
while abs( x( 1) -x( 0)) >p
x( 0) =x( 1)
x( 1) =( ( n -1.0) *x( 1) +A /x( 1)^( n -1.0)) /n
wend
NthRoot =x( 1)
end function
end
First estimate is: 1.071559602191682 ... and better is: 1.071547591944771 125'th root of 5642 by LB's exponentiation operator is 1.071547591944767 27^(1 / 3) 3.000000000000002 2^(1 / 2) 1.414213562374690 1024^(1 /10) 2.000000000000000
Lingo
on nthRoot (x, root)
return power(x, 1.0/root)
end
the floatPrecision = 8 -- only about display/string cast of floats
put nthRoot(4, 4)
-- 1.41421356
Logo
to about :a :b
output and [:a - :b < 1e-5] [:a - :b > -1e-5]
end
to root :n :a [:guess :a]
localmake "next ((:n-1) * :guess + :a / power :guess (:n-1)) / n
if about :guess :next [output :next]
output (root :n :a :next)
end
show root 5 34 ; 2.02439745849989
Lua
function nroot(root, num)
return num^(1/root)
end
M2000 Interpreter
Using stack statements PUSH, READ, OVER, SHIFT, DROP, NUMBER, FLUSH
Flush empty stack
Over 2 copy 2nd as new top (so 2nd now is 3rd)
Over 2,2 repeat Over 2 two times.
Shift 2 send top to 2nd, and 2nd to top (1st) (there is a SHFITBACK to revesre action)
Drop drop top
Number get top if is number, else raise error
Read, read a variable form top.
Functions parameters works with a read too
Function Root {
Read a, n%, d as double=1.e-4
......
}
because we can send any type and number if function, interpreter can make conversions if we declare that,
or if it not possible (no conversion done to a numeric variable if a string is in top of stack) we get an error.
Also if we send less values, and we didn't initialize variable before, we get error too.
Here we need to flush stack for other parameters if from an error anyone put more arguments.
(interpreter never count before call a user function, except for calling events by using event object,
so there there is a signature to follow)
n% is double inside.
Module Checkit {
Function Root (a, n%, d as double=1.e-4) {
if n%=0 then Error "Division by zero: 1/0"
if a<=0 then Error "Negative or zero number"
if n%=1 then = a : exit
Flush
n2=1-1/n%:a/=n%:n%--:Push a
{ Push 1: For i=1 to n% {Over 2 :Push Number*Number}
Over 2 : Push n2*Number + a/Number
Shift 2: Over 2, 2 :if Abs(Number-Number)>d Then loop
Drop
} Read a : = a
}
Print "square root single"
Print root(1.3346767~, 2, 1.e-9)
Print "square double"
Print root(1.3346767, 2, 1.e-9)
Print "square root decimal"
Print root(1.3346767@, 2, 1.e-9)
Print "internal square root, double"
Print 1.3346767^(1/2)
Print sqrt(1.3346767)
}
Checkit
Maple
The root command performs this task.
root(1728, 3);
root(1024, 10);
root(2.0, 2);
Output:
12
2
1.414213562
Mathematica
Root[A,n]
MATLAB
function answer = nthRoot(number,root)
format long
answer = number / root;
guess = number;
while not(guess == answer)
guess = answer;
answer = (1/root)*( ((root - 1)*guess) + ( number/(guess^(root - 1)) ) );
end
end
Sample Output:
nthRoot(2,2)
ans =
1.414213562373095
Maxima
nth_root(a, n) := block(
[x, y, d, p: fpprec],
fpprec: p + 10,
x: bfloat(a),
eps: 10.0b0^-p,
y: do (
d: bfloat((a / x^(n - 1) - x) / n),
if abs(d) < eps * x then return(x),
x: x + d
),
fpprec: p,
bfloat(y)
)$
Metafont
Metafont does not use IEEE floating point and we can't go beyond 0.0001 or it will loop forever.
vardef mnthroot(expr n, A) =
x0 := A / n;
m := n - 1;
forever:
x1 := (m*x0 + A/(x0 ** m)) / n;
exitif abs(x1 - x0) < abs(x0 * 0.0001);
x0 := x1;
endfor;
x1
enddef;
primarydef n nthroot A = mnthroot(n, A) enddef;
show 5 nthroot 34; % 2.0244
show 0.5 nthroot 7; % 49.00528
bye
=={{header|MK-61/52}}==
Instruction: ''number'' ^ ''degree'' В/О С/П
## NetRexx
{{trans|REXX}}
```netrexx
/*NetRexx program to calculate the Nth root of X, with DIGS accuracy. */
class nth_root
method main(args=String[]) static
if args.length < 2 then
do
say "at least 2 arguments expected"
exit
end
x = args[0]
root = args[1]
if args.length > 2 then digs = args[2]
if root=='' then root=2
if digs = null, digs = '' then digs=20
numeric digits digs
say ' x = ' x
say ' root = ' root
say 'digits = ' digs
say 'answer = ' root(x,root,digs)
method root(x,r,digs) static --procedure; parse arg x,R 1 oldR /*assign 2nd arg-->r and rOrig. */
/*this subroutine will use the */
/*digits from the calling prog. */
/*The default digits is 9. */
R = r
oldR = r
if r=0 then do
say
say '*** error! ***'
say "a root of zero can't be specified."
say
return '[n/a]'
end
R=R.abs() /*use absolute value of root. */
if x<0 & (R//2==0) then do
say
say '*** error! ***'
say "an even root can't be calculated for a" -
'negative number,'
say 'the result would be complex.'
say
return '[n/a]'
end
if x=0 | r=1 then return x/1 /*handle couple of special cases.*/
Rm1=R-1 /*just a fast version of ROOT-1 */
oldDigs=digs /*get the current number of digs.*/
dm=oldDigs+5 /*we need a little guard room. */
ax=x.abs() /*the absolute value of X. */
g=(ax+1)/r**r /*take a good stab at 1st guess. */
-- numeric fuzz 3 /*fuzz digits for higher roots. */
d=5 /*start with only five digits. */
/*each calc doubles precision. */
loop forever
d=d+d
if d>dm then d = dm /*double the digits, but not>DM. */
numeric digits d /*tell REXX to use D digits. */
old=0 /*assume some kind of old guess. */
loop forever
_=(Rm1*g**R+ax)/R/g**rm1 /*this is the nitty-gritty stuff.*/
if _=g | _=old then leave /*computed close to this before? */
old=g /*now, keep calculation for OLD. */
g=_ /*set calculation to guesstimate.*/
end
if d==dm then leave /*found the root for DM digits ? */
end
_=g*x.sign() /*correct the sign (maybe). */
if oldR<0 then return _=1/_ /*root < 0 ? Reciprocal it is.*/
numeric digits oldDigs /*re-instate the original digits.*/
return _/1 /*normalize the number to digs. */
NewLISP
(define (nth-root n a)
(let ((x1 a)
(x2 (div a n)))
(until (= x1 x2)
(setq x1 x2
x2 (div
(add
(mul x1 (- n 1))
(div a (pow x1 (- n 1))))
n)))
x2))
Nim
import math
proc nthroot(a, n): float =
var n = float(n)
result = a
var x = a / n
while abs(result-x) > 10e-15:
x = result
result = (1.0/n) * (((n-1)*x) + (a / pow(x, n-1)))
echo nthroot(34.0, 5)
echo nthroot(42.0, 10)
echo nthroot(5.0, 2)
Output:
2.0243974584998852e+00
1.4531984602822678e+00
2.2360679774997898e+00
Objeck
{{trans|C}}
class NthRoot {
function : Main(args : String[]) ~ Nil {
NthRoot(5, 34, .001)->PrintLine();
}
function : NthRoot(n : Int, A: Float, p : Float) ~ Float {
x := Float->New[2];
x[0] := A;
x[1] := A / n;
while((x[1] - x[0])->Abs() > p) {
x[0] := x[1];
x[1] := ((n - 1.0) * x[1] + A / x[1]->Power(n - 1.0)) / n;
};
return x[1];
}
}
OCaml
{{trans|C}}
let nthroot ~n ~a ?(tol=0.001) () =
let nf = float n in let nf1 = nf -. 1.0 in
let rec iter x =
let x' = (nf1 *. x +. a /. (x ** nf1)) /. nf in
if tol > abs_float (x -. x') then x' else iter x' in
iter 1.0
;;
let () =
Printf.printf "%g\n" (nthroot 10 (7131.5 ** 10.0) ());
Printf.printf "%g\n" (nthroot 5 34.0 ());
;;
Octave
Octave has it's how nthroot function.
r = A.^(1./n)
Here it is another implementation (after Tcl)
{{trans|Tcl}}
function r = m_nthroot(n, A)
x0 = A / n;
m = n - 1;
while(1)
x1 = (m*x0 + A./ x0 .^ m) / n;
if ( abs(x1-x0) < abs(x0 * 1e-9) )
r = x1;
return
endif
x0 = x1;
endwhile
endfunction
Here is an more elegant way by computing the successive differences in an explicit way:
function r = m_nthroot(n, A)
r = A / n;
m = n - 1;
do
d = (A ./ r .^ m - r) / n;
r+= d;
until (abs(d) < abs(r * 1e-9))
endfunction
Show its usage and the built-in nthroot function
m_nthroot(10, 7131.5 .^ 10)
nthroot(7131.5 .^ 10, 10)
m_nthroot(5, 34)
nthroot(34, 5)
m_nthroot(0.5, 7)
nthroot(7, .5)
Oforth
Float method: nthroot(n)
1.0 doWhile: [ self over n 1 - pow / over - n / tuck + swap 0.0 <> ] ;
{{out}}
734 10.0 powf nthroot(10) println
734
2.0 nthroot(2) println
1.41421356237309
34.0 nthroot(5) println
2.02439745849989
Oz
declare
fun {NthRoot NInt A}
N = {Int.toFloat NInt}
fun {Next X}
( (N-1.0)*X + A / {Pow X N-1.0} ) / N
end
in
{Until Value.'==' Next A/N}
end
fun {Until P F X}
case {F X}
of NX andthen {P NX X} then X
[] NX then {Until P F NX}
end
end
in
{Show {NthRoot 2 2.0}}
PARI/GP
root(n,A)=A^(1/n);
Pascal
See [[Nth_root#Delphi | Delphi]]
Perl
{{trans|Tcl}}
use strict;
sub nthroot ($$)
{
my ( $n, $A ) = @_;
my $x0 = $A / $n;
my $m = $n - 1.0;
while(1) {
my $x1 = ($m * $x0 + $A / ($x0 ** $m)) / $n;
return $x1 if abs($x1 - $x0) < abs($x0 * 1e-9);
$x0 = $x1;
}
}
print nthroot(5, 34), "\n";
print nthroot(10, 7131.5 ** 10), "\n";
print nthroot(0.5, 7), "\n";
Perl 6
sub nth-root ($n, $A, $p=1e-9)
{
my $x0 = $A / $n;
loop {
my $x1 = (($n-1) * $x0 + $A / ($x0 ** ($n-1))) / $n;
return $x1 if abs($x1-$x0) < abs($x0 * $p);
$x0 = $x1;
}
}
say nth-root(3,8);
Phix
{{trans|AWK}} (main loop) {{trans|C}} (use of pow_ instead of power)
function pow_(atom x, integer e)
atom r = 1
for i=1 to e do
r *= x
end for
return r
end function
function nth_root(atom y,n)
atom eps = 1e-15 -- relative accuracy
atom x = 1
while 1 do
-- atom d = ( y / power(x,n-1) - x ) / n
atom d = ( y / pow_(x,n-1) - x ) / n
x += d
atom e = eps*x -- absolute accuracy
if d > -e and d < e then exit end if
end while
return {y,n,x,power(y,1/n)}
end function
?nth_root(1024,10)
?nth_root(27,3)
?nth_root(2,2)
?nth_root(5642,125)
--?nth_root(7,0.5) -- needs power(), not pow_()
?nth_root(4913,3)
?nth_root(8,3)
?nth_root(16,2)
?nth_root(16,4)
?nth_root(125,3)
?nth_root(1000000000,3)
?nth_root(1000000000,9)
{{out}} Shows inputs and both the iterative and builtin results.
{1024,10,2,2}
{27,3,3,3}
{2,2,1.414213562,1.414213562}
{5642,125,1.071547592,1.071547592}
{4913,3,17,17.0}
{8,3,2,2}
{16,2,4,4}
{16,4,2,2}
{125,3,5,5}
{1000000000,3,1000,1000.0}
{1000000000,9,10,10.0}
PHP
function nthroot($number, $root, $p = P)
{
$x[0] = $number;
$x[1] = $number/$root;
while(abs($x[1]-$x[0]) > $p)
{
$x[0] = $x[1];
$x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root;
}
return $x[1];
}
PicoLisp
(load "@lib/math.l")
(de nthRoot (N A)
(let (X1 A X2 (*/ A N))
(until (= X1 X2)
(setq
X1 X2
X2 (*/
(+
(* X1 (dec N))
(*/ A 1.0 (pow X1 (* (dec N) 1.0))) )
N ) ) )
X2 ) )
(prinl (format (nthRoot 2 2.0) *Scl))
(prinl (format (nthRoot 3 12.3) *Scl))
(prinl (format (nthRoot 4 45.6) *Scl))
Output:
1.414214
2.308350
2.598611
PL/I
/* Finds the N-th root of the number A */
root: procedure (A, N) returns (float);
declare A float, N fixed binary;
declare (xi, xip1) float;
xi = 1; /* An initial guess */
do forever;
xip1 = ((n-1)*xi + A/xi**(n-1) ) / n;
if abs(xip1-xi) < 1e-5 then leave;
xi = xip1;
end;
return (xi);
end root;
Results:
The 2-th root of 4.00000E+0000 is 2.00000E+0000
The 5-th root of 3.20000E+0001 is 2.00000E+0000
The 3-th root of 2.70000E+0001 is 3.00000E+0000
The 2-th root of 2.00000E+0000 is 1.41422E+0000
The 3-th root of 1.00000E+0002 is 4.64159E+0000
PowerShell
This sample implementation does not use [System.Math] classes.
#NoTeS: This sample code does not validate inputs
# Thus, if there are errors the 'scary' red-text
# error messages will appear.
#
# This code will not work properly in floating point values of n,
# and negative values of A.
#
# Supports negative values of n by reciprocating the root.
$epsilon=1E-10 #Sample Epsilon (Precision)
function power($x,$e){ #As I said in the comment
$ret=1
for($i=1;$i -le $e;$i++){
$ret*=$x
}
return $ret
}
function root($y,$n){ #The main Function
if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} #This checks if n is negative.
$ans=1
do{
$d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp
$ans+=$d
} while ($d -lt -$epsilon -or $d -gt $epsilon)
if (0+$n -lt 0){return 1/$ans} else {return $ans}
}
#Sample Inputs
root 625 2
root 2401 4
root 2 -2
root 1.23456789E-20 34
root 9.87654321E20 10 #Quite slow here, I admit...
((root 5 2)+1)/2 #Extra: Computes the golden ratio
((root 5 2)-1)/2
{{Out}}
PS> .\NTH.PS1
25
7
0.707106781186548
0.259690655650288
125.736248016373
1.61803398874989
0.618033988749895
PS>
PureBasic
#Def_p=0.001
Procedure.d Nth_root(n.i, A.d, p.d=#Def_p)
Protected Dim x.d(1)
x(0)=A: x(1)=A/n
While Abs(x(1)-x(0))>p
x(0)=x(1)
x(1)=((n-1.0)*x(1)+A/Pow(x(1),n-1.0))/n
Wend
ProcedureReturn x(1)
EndProcedure
;//////////////////////////////
Debug "125'th root of 5642 is"
Debug Pow(5642,1/125)
Debug "First estimate is:"
Debug Nth_root(125,5642)
Debug "And better:"
Debug Nth_root(125,5642,0.00001)
'''Outputs 125'th root of 5642 is 1.0715475919447675 First estimate is: 1.0715596021916822 And better: 1.0715475919447714
Python
from decimal import Decimal, getcontext
def nthroot (n, A, precision):
getcontext().prec = precision
n = Decimal(n)
x_0 = A / n #step 1: make a while guess.
x_1 = 1 #need it to exist before step 2
while True:
#step 2:
x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1))))
if x_0 == x_1:
return x_1
print nthroot(5, 34, 10)
print nthroot(10,42, 20)
print nthroot(2, 5, 400)
R
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps))
{
ifelse(A < 1, x0 <- A * n, x0 <- A / n)
repeat
{
x1 <- ((n-1)*x0 + A / x0^(n-1))/n
if(abs(x1 - x0) > tol) x0 <- x1 else break
}
x1
}
nthroot(7131.5^10, 10) # 7131.5
nthroot(7, 0.5) # 49
Racket
#lang racket
(define (nth-root number root (tolerance 0.001))
(define (acceptable? next current)
(< (abs (- next current)) tolerance))
(define (improve current)
(/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root))
(define (loop current)
(define next-guess (improve current))
(if (acceptable? next-guess current)
next-guess
(loop next-guess)))
(loop 1.0))
REXX
/*REXX program calculates the Nth root of X, with DIGS (decimal digits) accuracy. */
parse arg x root digs . /*obtain optional arguments from the CL*/
if x=='' | x=="," then x= 2 /*Not specified? Then use the default.*/
if root=='' | root=="," then root= 2 /* " " " " " " */
if digs=='' | digs=="," then digs=65 /* " " " " " " */
numeric digits digs /*set the decimal digits to DIGS. */
say ' x = ' x /*echo the value of X. */
say ' root = ' root /* " " " " ROOT. */
say ' digits = ' digs /* " " " " DIGS. */
say ' answer = ' root(x, root) /*show the value of ANSWER. */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
root: procedure; parse arg x 1 Ox, r 1 Or /*arg1 ──► x & Ox, 2nd ──► r & Or*/
if r=='' then r=2 /*Was root specified? Assume √. */
if r=0 then return '[n/a]' /*oops-ay! Can't do zeroth root.*/
complex= x<0 & R//2==0 /*will the result be complex? */
oDigs=digits() /*get the current number of digs.*/
if x=0 | r=1 then return x/1 /*handle couple of special cases.*/
dm=oDigs+5 /*we need a little guard room. */
r=abs(r); x=abs(x) /*the absolute values of R and X.*/
rm=r-1 /*just a fast version of ROOT -1*/
numeric form /*take a good guess at the root─┐*/
parse value format(x,2,1,,0) 'E0' with ? 'E' _ . /* ◄────────────────────────────┘*/
g= (? / r'E'_ % r) + (x>1) /*kinda uses a crude "logarithm".*/
d=5 /*start with five decimal digits.*/
do until d==dm; d=min(d+d,dm) /*each time, precision doubles. */
numeric digits d /*tell REXX to use D digits. */
old=-1 /*assume some kind of old guess. */
do until old=g; old=g /*where da rubber meets da road─┐*/
g=format((rm*g**r+x)/r/g**rm,, d-2) /* ◄────── the root computation─┘*/
end /*until old=g*/ /*maybe until the cows come home.*/
end /*until d==dm*/ /*and wait for more cows to come.*/
if g=0 then return 0 /*in case the jillionth root = 0.*/
if Or<0 then g=1/g /*root < 0 ? Reciprocal it is! */
if \complex then g=g*sign(Ox) /*adjust the sign (maybe). */
numeric digits oDigs /*reinstate the original digits. */
return (g/1) || left('j', complex) /*normalize # to digs, append j ?*/
'''output''' when using the default inputs:
x = 2
root = 2
digits = 65
answer = 1.414213562373095048801688724209698078569671875376948073176679738
'''output''' when using for input: 10 3
x = 10
root = 3
digits = 65
answer = 2.1544346900318837217592935665193504952593449421921085824892355063
'''output''' when using for input: 625 -4
x = 625
root = -4
digits = 65
answer = 0.2
'''output''' when using for input: 100.666 47
x = 100.666
root = 47
digits = 65
answer = 1.1030990940616109102886569991014966919115206420386192403152621652
'''output''' when using for input: -256 8
x = -256
root = 8
digits = 65
answer = 2j
'''output''' when using for input: 12345678900098765432100.00987654321000123456789e333 19
x = 12345678900098765432100.00987654321000123456789e333
root = 19
digits = 65
answer = 4886828567991886455.3257854108687610458584138783288904955196401434
Ring
decimals(12)
see "cube root of 5 is : " + root(3, 5, 0) + nl
func root n, a, d
y = 0 x = a / n
while fabs (x - y) > d
y = ((n - 1)*x + a/pow(x,(n-1))) / n
temp = x
x = y
y = temp
end
return x
Output:
cube root of 5 is : 1.709975946677
Ruby
def nthroot(n, a, precision = 1e-5)
x = Float(a)
begin
prev = x
x = ((n - 1) * prev + a / (prev ** (n - 1))) / n
end while (prev - x).abs > precision
x
end
p nthroot(5,34) # => 2.02439745849989
Run BASIC
print "Root 125th Root of 5643 Precision .001 ";using( "#.###############", NthRoot( 125, 5642, 0.001 ))
print "125th Root of 5643 Precision .001 ";using( "#.###############", NthRoot( 125, 5642, 0.001 ))
print "125th Root of 5643 Precision .00001 ";using( "#.###############", NthRoot( 125, 5642, 0.00001))
print " 3rd Root of 27 Precision .00001 ";using( "#.###############", NthRoot( 3, 27, 0.00001))
print " 2nd Root of 2 Precision .00001 ";using( "#.###############", NthRoot( 2, 2, 0.00001))
print " 10th Root of 1024 Precision .00001 ";using( "#.###############", NthRoot( 10, 1024, 0.00001))
wait
function NthRoot( root, A, precision)
x0 = A
x1 = A /root
while abs( x1 -x0) >precision
x0 = x1
x1 = x1 / 1.0 ' force float
x1 = (( root -1.0) *x1 +A /x1^( root -1.0)) /root
wend
NthRoot =x1
end function
end
125th Root of 5643 Precision .001 1.071559602456735
125th Root of 5643 Precision .00001 1.071547591944771
3rd Root of 27 Precision .00001 3.000000000000001
2nd Root of 2 Precision .00001 1.414213562374690
10th Root of 1024 Precision .00001 2.000000000000000
Sather
{{trans|Octave}}
class MATH is
nthroot(n:INT, a:FLT):FLT
pre n > 0
is
x0 ::= a / n.flt;
m ::= n - 1;
loop
x1 ::= (m.flt * x0 + a/(x0^(m.flt))) / n.flt;
if (x1 - x0).abs < (x0 * 1.0e-9).abs then
return x1;
end;
x0 := x1;
end;
end;
end;
class MAIN is
main is
a:FLT := 2.5 ^ 10.0;
#OUT + MATH::nthroot(10, a) + "\n";
end;
end;
Scala
Using tail recursion:
def nroot(n: Int, a: Double): Double = {
@tailrec
def rec(x0: Double) : Double = {
val x1 = ((n - 1) * x0 + a/math.pow(x0, n-1))/n
if (x0 <= x1) x0 else rec(x1)
}
rec(a)
}
Alternatively, you can implement the iteration with an iterator like so:
def fallPrefix(itr: Iterator[Double]): Iterator[Double] = itr.sliding(2).dropWhile(p => p(0) > p(1)).map(_.head)
def nrootLazy(n: Int)(a: Double): Double = fallPrefix(Iterator.iterate(a){r => (((n - 1)*r) + (a/math.pow(r, n - 1)))/n}).next
Scheme
(define (root number degree tolerance)
(define (good-enough? next guess)
(< (abs (- next guess)) tolerance))
(define (improve guess)
(/ (+ (* (- degree 1) guess) (/ number (expt guess (- degree 1)))) degree))
(define (*root guess)
(let ((next (improve guess)))
(if (good-enough? next guess)
guess
(*root next))))
(*root 1.0))
(display (root (expt 2 10) 10 0.1))
(newline)
(display (root (expt 2 10) 10 0.01))
(newline)
(display (root (expt 2 10) 10 0.001))
(newline)
Output: 2.04732932236839 2.00463204835482 2.00004786858167
Seed7
The nth root of the number 'a' can be computed with the exponentiation operator: 'a ** (1 / n)'. An alternate function which uses Newton's method is:
const func float: nthRoot (in integer: n, in float: a) is func
result
var float: x1 is 0.0;
local
var float: x0 is 0.0;
begin
x0 := a;
x1 := a / flt(n);
while abs(x1 - x0) >= abs(x0 * 1.0E-9) do
x0 := x1;
x1 := (flt(pred(n)) * x0 + a / x0 ** pred(n)) / flt(n);
end while;
end func;
Original source: [http://seed7.sourceforge.net/algorith/math.htm#nthRoot]
Sidef
{{trans|Ruby}}
func nthroot(n, a, precision=1e-5) {
var x = 1.float
var prev = 0.float
while ((prev-x).abs > precision) {
prev = x;
x = (((n-1)*prev + a/(prev**(n-1))) / n)
}
return x
}
say nthroot(5, 34) # => 2.024397458501034082599817835297912829678314204
A minor optimization would be to calculate the successive ''int(n-1)'' square roots of a number, then raise the result to the power of ''2**(int(n-1) / n)''.
func nthroot_fast(n, a, precision=1e-5) {
{ a = nthroot(2, a, precision) } * int(n-1)
a ** (2**int(n-1) / n)
}
say nthroot_fast(5, 34, 1e-64) # => 2.02439745849988504251081724554193741911462170107
Smalltalk
{{works with|GNU Smalltalk}}
{{trans|Tcl}}
Number extend [
nthRoot: n [
|x0 m x1|
x0 := (self / n) asFloatD.
m := n - 1.
[true] whileTrue: [
x1 := ( (m * x0) + (self/(x0 raisedTo: m))) / n.
((x1 - x0) abs) < ((x0 * 1e-9) abs)
ifTrue: [ ^ x1 ].
x0 := x1
]
]
].
(34 nthRoot: 5) displayNl.
((7131.5 raisedTo: 10) nthRoot: 10) displayNl.
(7 nthRoot: 0.5) displayNl.
SPL
nthr(n,r) <= n^(1/r)
nthroot(n,r)=
a = n/r
g = n
> g!=a
g = a
a = (1/r)*(((r-1)*g)+(n/(g^(r-1))))
<
<= a
.
#.output(nthr(2,2))
#.output(nthroot(2,2))
{{out}}
1.4142135623731
1.41421356237309
Tcl
The easiest way is to just use the pow function (or exponentiation operator) like this:
proc nthroot {n A} {
expr {pow($A, 1.0/$n)}
}
However that's hardly tackling the problem itself. So here's how to do it using Newton-Raphson and a self-tuning termination test.
{{works with|Tcl|8.5}}
proc nthroot {n A} {
set x0 [expr {$A / double($n)}]
set m [expr {$n - 1.0}]
while 1 {
set x1 [expr {($m*$x0 + $A/$x0**$m) / $n}]
if {abs($x1 - $x0) < abs($x0 * 1e-9)} {
return $x1
}
set x0 $x1
}
}
Demo:
puts [nthroot 2 2]
puts [nthroot 5 34]
puts [nthroot 5 [expr {34**5}]]
puts [nthroot 10 [expr 7131.5**10]]
puts [nthroot 0.5 7]; # Squaring!
Output:
1.414213562373095
2.0243974584998847
34.0
7131.5
49.0
Ursala
The nthroot function defined below takes a natural number n to the function that returns the n-th root of its floating point argument. Error is on the order of machine precision because the stopping criterion is either a fixed point or a repeating cycle.
#import nat
#import flo
nthroot =
-+
("n","n-1"). "A". ("x". div\"n" plus/times("n-1","x") div("A",pow("x","n-1")))^== 1.,
float^~/~& predecessor+-
This implementation is unnecessary in practice due to the availability of the library function pow, which performs exponentiation and allows fractional exponents. Here is a test program.
#cast %eL
examples =
<
nthroot2 2.,
nthroot5 34.,
nthroot5 pow(34.,5.),
nthroot10 pow(7131.5,10.)>
output:
<
1.414214e+00,
2.024397e+00,
3.400000e+01,
7.131500e+03>
VBA
{{trans|Phix}} The internal power operator "^" is used in stead of an auxiliary pow_ function and the accuracy has been reduced.
Private Function nth_root(y As Double, n As Double)
Dim eps As Double: eps = 0.00000000000001 '-- relative accuracy
Dim x As Variant: x = 1
Do While True
d = (y / x ^ (n - 1) - x) / n
x = x + d
e = eps * x '-- absolute accuracy
If d > -e And d < e Then
Exit Do
End If
Loop
Debug.Print y; n; x; y ^ (1 / n)
End Function
Public Sub main()
nth_root 1024, 10
nth_root 27, 3
nth_root 2, 2
nth_root 5642, 125
nth_root 7, 0.5
nth_root 4913, 3
nth_root 8, 3
nth_root 16, 2
nth_root 16, 4
nth_root 125, 3
nth_root 1000000000, 3
nth_root 1000000000, 9
End Sub
{{out}}
1024 10 2 2
27 3 3 3
2 2 1,41421356237309 1,4142135623731
5642 125 1,07154759194477 1,07154759194477
7 0,5 49 49
4913 3 17 17
8 3 2 2
16 2 4 4
16 4 2 2
125 3 5 5
1000000000 3 1000 1000
1000000000 9 10 10
XPL0
include c:\cxpl\stdlib;
func real NRoot(A, N); \Return the Nth root of A
real A, N;
real X, X0, Y;
int I;
[X:= 1.0; \initial guess
repeat X0:= X;
Y:= 1.0;
for I:= 1 to fix(N)-1 do Y:= Y*X0;
X:= ((N-1.0)*X0 + A/Y) / N;
until abs(X-X0) < 1.0E-15; \(until X=X0 doesn't always work)
return X;
];
[Format(5, 15);
RlOut(0, NRoot( 2., 2.)); CrLf(0);
RlOut(0, Power( 2., 0.5)); CrLf(0); \for comparison
RlOut(0, NRoot(27., 3.)); CrLf(0);
RlOut(0, NRoot(1024.,10.)); CrLf(0);
]
Output:
1.414213562373100
1.414213562373100
3.000000000000000
2.000000000000000
Yabasic
{{trans|AWK}}
data 10, 1024, 3, 27, 2, 2, 125, 5642, 4, 16, 0, 0
do
read e, b
if e = 0 break
print "The ", e, "th root of ", b, " is ", b^(1/e), " (", nthroot(b, e), ")"
loop
sub nthroot(y, n)
local eps, x, d, e
eps = 1e-15 // relative accuracy
x = 1
repeat
d = ( y / ( x^(n-1) ) - x ) / n
x = x + d
e = eps * x // absolute accuracy
until(not(d < -e or d > e ))
return x
end sub
zkl
{{trans|Ruby}}
fcn nthroot(nth,a,precision=1.0e-5){
x:=prev:=a=a.toFloat(); n1:=nth-1;
do{
prev=x;
x=( prev*n1 + a/prev.pow(n1) ) / nth;
}
while( not prev.closeTo(x,precision) );
x
}
nthroot(5,34) : "%.20f".fmt(_).println() # => 2.02439745849988828041
{{omit from|M4}}